CHEMISTRY COMMUNITY

Christopher Phung 3G

Here is the picture on the problem that I'm stuck on.

See that there is a curved arrow from the hydrogen bond to the C-O bond to make a C=O bond.
Would I be wrong to draw the curved arrow from the hydrogen bond to the C, and then another curved arrow from the C to the C-O bond to make the C=O bond?

Christopher Phung 3G

Instead of drawing a curved arrow from one bond (one that's being broken) to another bond (to make a pi bond), would I be wrong to draw two curved arrows - one going from the bond that's being broken to the atom, and a lone pair from the atom to the other bond to make a pi bond? I just don't feel ri...

Christopher Phung 3G

I'm a little lost when it comes to drawing the curved arrows in a reaction mechanism. I was working on Question 6 from the 2014 Final and I was confused with where to begin the curved arrow and where to end it. For example, for one of the steps, I drew a curved arrow going from a bond to an atom, bu...

Christopher Phung 3G

I'm really lost in all this. I don't recall anything about priority! How are cis and trans isomers named? I get that trans isomers are on opposite sides, and that cis are on the same side, but there seems to be more to it than just that. I was working on this same problem and didn't think about whet...

Christopher Phung 3G

Ah, if only I had done the homework! Hahaha. Just for future questions, I approached the first problem that you listed using the Henderson-Hasselbalch equation and arrived at a different answer. Now that I know it's wrong, I've been thinking about when it is appropriate to use the H-H equation. What...

Christopher Phung 3G

Great explanation! Your work is easy to follow and makes a lot of sense. The way I approached this question was a bit less orthodox - I began by supposing the initial concentration of acetic acid was 1.0 x 10 -7 M. Using its Ka value, I found the equilibrium concentration of H + to be much less than...

Christopher Phung 3G

That's a good question. Maybe it's an error in the book. I'm curious what Chem_Mod has to say about this.

Christopher Phung 3G

So to solve the final pH for 12.23, we need to find the pH of 0.005 M HCl. HCl dissociates completely, so for every liter of 0.005 M HCl, we have 0.005 moles of H⁺. That is, we have 0.005 M H⁺. The pH of 0.005 M H⁺ is 2.3.

Christopher Phung 3G

The initial pH of the titration is the pH of the sample. If the problem you're working on is 12.23, then the quick and dirty way of finding the pH at the stoichiometric point is to recognize that both the sample and titrant are strong acids and bases, which means the stoichiometric point is at 7 pH....

Christopher Phung 3G

An example of a Lewis acid is AlH3 (aluminum hydride). Notice how the central aluminum atom doesn't have its octet satisfied, which means it has a tendency of stripping a lone pair off of another species, rendering it a Lewis acid (electron recipient/taker). An example of a Bronsted acid is NH4 + (a...

Christopher Phung 3G

Would you happen to have the Ka of CH3COOH and NaCH3CO2?

Christopher Phung 3G

Is there any experimental data or additional information related to this question, such as the pH or concentration of hydronium at equilibrium? The constant K is a fraction of the concentrations of products over the concentrations of reactants at equilibrium. Without any information on the concentra...

Christopher Phung 3G

If you're not doing a titration or a dilution, there isn't really an "initial" or "final" pH. The pH is a number describing the concentration of hydronium ions in solution at equilibrium. If you are doing a titration or dilution, then both the initial volume and final volumes mus...

Christopher Phung 3G

Part a: You are right - the process needs to be done in reverse. However, the ICE box still remains. We have the following ICE box with the given information: \begin{matrix} & HClO_{2} \rightleftharpoons & H^{+} + ClO_{2}^{-} \\ I & 0.1 & 0 & 0 \\ C & -x & +x & +x \\ ...

Christopher Phung 3G

First, we should write out the net ionic equation. The first step in that is to write out everything that can be ionized into ions. This yields the following: NH 4 + + I - + K + + NH 2 - --> K + + I - + 2NH 3 (liquid) (all species except NH 3 are in ammonia solvent) Next, we cross out all ionic spec...

Christopher Phung 3G

For part b: Recall that pOH = -log[ - OH] where log stands for "common log." Isolating [ - OH] on the right side of the equation, we can find that it equals 10 -pOH . Since we know pOH from part a (14 - pH = pOH = 0.75), to solve for [ - OH] we calculate 10 -0.75 which is approx. 0.18. Fo...

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